How do I complete this "Cayley table" or binary operation table? I have an algebraic structure $(S,\cdot)$ and $a,b,c,d \in S$ where $a,b,c,d$ are not necessarily four distinct elements. This is part of a larger problem that I am working on and based on what I know, the following is the operation table or "Cayley table" (quotes because this isn't necessarily a group):
$$\begin{array}{c|cccc}
\cdot & a & b & c & d \\\hline
a & a &   & c & \\
b &   & b &   & d \\
c &   & c &   & a \\
d & d &   & b &   \\ 
\end{array}
$$
The table is made such that $d \cdot c = b$.
The minimum assumption I want to make is that one of associativity, because I can't even manipulate equations without it.
My primary question is: Is it possible to fill in any more spaces on this binary operation table with the information given and assuming associativity only?
If not, what (minimum number of) assumptions would I have to make to fill up the table completely? 
 A: If $a=b=c=d$, then any possible filling is trivially associative. Therefore, assumption of associativity alone can't rule out any filling (or force any), so you can't say anything about permissible fillings assuming just associativity alone, and no assumptions (about values of specific products) short of assuming all of them will be sufficient.
I think the problem (highlighted by the previous paragraph) is that you are not asking the right question. The right question would be, what kind of associative operations can satisfy this partial Cayley table.
Under some assumptions about (in)equality of various of $a,b,c,d$ you can find a unique correct filling up to obvious identification, or equivalently, a unique binary operation satisfying the table. 
For example, if you assume that $a=b=c=d$, then $\cdot$ is just the trivial group operation, and if you assume that $a=b\neq c=d$, then $\cdot$ is the two-element group operation with neutral element $a=b$. In fact, for those two, you don't even need to assume associativity, as it follows from the partial table.
A: Computer search finds these three completions and no others:
$$\begin{array}{c|cccc}
\cdot & a & b & c & d \\\hline
a & a & c & c & a \\
b & d & b & b & d \\
c & a & c & c & a \\
d & d & b & b & d \\ 
\end{array}
$$
$$\begin{array}{c|cccc}
\cdot & a & b & c & d \\\hline
a & a & a & c & c \\
b & b & b & d & d \\
c & c & c & a & a \\
d & d & d & b & b \\ 
\end{array}
$$
$$\begin{array}{c|cccc}
\cdot & a & b & c & d \\\hline
a & a & b & c & d \\
b & a & b & c & d \\
c & d & c & b & a \\
d & d & c & b & a \\ 
\end{array}
$$
